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College of San Mateo

# Differentiate the function.$h(x) = \ln (x + \sqrt {x^2 - 1})$

## h'(x)= 1/(x + sqrt(x^2-1)) * (1 + 0.5*(x^2 -1)^(-0.5)) * (2x)

Derivatives

Differentiation

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##### Heather Z.

Oregon State University

##### Kristen K.

University of Michigan - Ann Arbor

##### Samuel H.

University of Nottingham

Lectures

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### Video Transcript

Okay, let's look at this problem, associate is a long resume function. The first step we want to do is to use that differential rule of natural log function. So we got h its prime X equals two. One over X plus square root. Oh X squared minus one. Okay, we treat this as a whole and then we have to differentiate this thing. Then we got X plus square root X squared minus one crime. Right? How do we deal with saying? Well, we have to use the sum rule and chain rule as well as the power route to solve this. And for before that I want to rewrite this as a power for. So we have one over X plus Beirut X squared minus one times X plus on half mm I'm plus square minus one. It was a one half power and then prime. Okay then copy this down here. X square plus square square root tech square minus one climbs. Okay, so the dream team of actually just one right And the duty of of this thing. We use the power route we got x squared minus one times. Okay, Just took one minus one half in just 91 half. Right, okay. Then we have to use the chain group because this is another single variable. It's a expressive expression. Right? So we have to you will find the derivative of X squared minus one. Just the two X. Okay then coming this down here. What over X plus square root square minus one times. Okay, for this whole thing, the negative, what has of X square minus one is just one over square root X squared minus one. Right? Then we can write this as square root of square minus one. The numerator, it's one times two X. It's just two X. Okay? Yeah, then we can write down yes one over X plus square root X squared minus one. And we rewrite this as Okay? So first we can we see this to to we can cancel all the tooth and we write this as x squared minus one over x squared minus one. Right? That just you close to one, right? And we plus X over square minus one. Okay, so here we got copy this down. As usual. We got X square minus one square root of X squared minus one plus X over x squared minus one square root of x squared minus one. Right. And we noticed that this thing on the numerous not the denominator and this thing on the numerator, they're actually the same. Right? So we can cancel out these two items. And finally all the things left is just one over square root of X. Square one. Is what? Okay, that's a solution of this problem. Thank you.

College of San Mateo

#### Topics

Derivatives

Differentiation

##### Heather Z.

Oregon State University

##### Kristen K.

University of Michigan - Ann Arbor

##### Samuel H.

University of Nottingham

Lectures

Join Bootcamp